1 SPECTRUM OF A CONTINUOUS OPERATOR 313
in Chapter XV, and its most important applications in Chapters XXI (repre-
sentations of compact groups), XXII (harmonic analysis), and XXIII (linear
functional equations).
1. SPECTRUM OF A CONTINUOUS OPERATOR
Let E be a complex normed space; a linear mapping u of E into itself is
often called an operator in E. The set &(E; E) of continuous operators (which
we will write simply «£?(£)) is a complex normed space (Section 5.7); it is also
a noncommutative algebra over C, the "product'* being the mapping
(u, v)-*uav, also written (u, v) -> uv. The identity mapping of E is the unit
element of 3? (E), written 1E . The mappings (u, v)-+u + v and (u,v)~+uov
are continuous in jSf(E) x jSf (E) (5.7.5).
We say that a complex number C is a regular value for a continuous
operator u if u - C ' 1 E has an inverse v^ in &(E) (i.e. is a linear homeomorphism
of E onto itself). The complex numbers C which are not regular for u are called
spectral values of u and the set of spectral values of u is called the spectrum
sp(w) of u.
If C e C is such that the kernel of u - C • 1E *s not reduced to 0, then £
is a spectral value of u; such spectral values are called eigenvalues of u; any
vector x^Q in the kernel of u - C • 1E> i-e- such that w(jc) = fjc, is called an
eigenvector of w corresponding to the eigenvalue C; these eigenvectors and 0
form a closed vector subspace of E, the kernel of u — C • 1E » also called the
eigenspace of u corresponding to the eigenvalue C, and written E(Q or
When E has ym/te dimension /?, elementary linear algebra shows that
any spectral value of an operator u is an eigenvalue of u (A.4.19); the
spectrum of u is a finite set of at most n elements, which are the roots of the
characteristic polynomial det(w — C * IE) °f w» of degree n (A.6.9). But if E
is infinite dimensional, there may exist spectral values which are not
eigenvalues.
Example
(11.1.1) Let E be a complex Hilbert space, (an)n%i a total orthonormal
system in E (6.6.1). To each vector x = ££„*« in E (with W2 = £IU2) we
n n
associate the vector u(x) = ]T („ an+l ; it is readily verified that u is linear and
n
\\u(x)\\ = \\x\\, hence (5.5.1) u is continuous. Moreover, u(E) is the subspace of
E orthogonal to al9 hence u is not surjective, and this shows that C = 0 is a
spectral value of u; but u(x) = 0 implies x = 0, hence 0 is not an eigenvalue
of u.